Integrand size = 23, antiderivative size = 405 \[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^2} \, dx=-\frac {2 b e \operatorname {AppellF1}\left (1-m,\frac {1-m}{2},\frac {1-m}{2},2-m,-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right ) \left (-\frac {a (1-\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} \left (\frac {a (1+\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}}{a^3 d (1-m)}+\frac {b^2 e \operatorname {AppellF1}\left (2-m,\frac {1-m}{2},\frac {1-m}{2},3-m,-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right ) \left (-\frac {a (1-\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} \left (\frac {a (1+\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}}{a^3 d (2-m) (b+a \cos (c+d x))}+\frac {\cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{a^2 d e (1+m) \sqrt {\cos ^2(c+d x)}} \]
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Time = 0.52 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3957, 2991, 2722, 2782} \[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^2} \, dx=\frac {b^2 e (e \sin (c+d x))^{m-1} \left (-\frac {a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \left (\frac {a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \operatorname {AppellF1}\left (2-m,\frac {1-m}{2},\frac {1-m}{2},3-m,-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right )}{a^3 d (2-m) (a \cos (c+d x)+b)}-\frac {2 b e (e \sin (c+d x))^{m-1} \left (-\frac {a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \left (\frac {a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \operatorname {AppellF1}\left (1-m,\frac {1-m}{2},\frac {1-m}{2},2-m,-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right )}{a^3 d (1-m)}+\frac {\cos (c+d x) (e \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\sin ^2(c+d x)\right )}{a^2 d e (m+1) \sqrt {\cos ^2(c+d x)}} \]
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Rule 2722
Rule 2782
Rule 2991
Rule 3957
Rubi steps \begin{align*} \text {integral}& = \int \frac {\cos ^2(c+d x) (e \sin (c+d x))^m}{(-b-a \cos (c+d x))^2} \, dx \\ & = \int \left (\frac {(e \sin (c+d x))^m}{a^2}+\frac {b^2 (e \sin (c+d x))^m}{a^2 (b+a \cos (c+d x))^2}-\frac {2 b (e \sin (c+d x))^m}{a^2 (b+a \cos (c+d x))}\right ) \, dx \\ & = \frac {\int (e \sin (c+d x))^m \, dx}{a^2}-\frac {(2 b) \int \frac {(e \sin (c+d x))^m}{b+a \cos (c+d x)} \, dx}{a^2}+\frac {b^2 \int \frac {(e \sin (c+d x))^m}{(b+a \cos (c+d x))^2} \, dx}{a^2} \\ & = -\frac {2 b e \operatorname {AppellF1}\left (1-m,\frac {1-m}{2},\frac {1-m}{2},2-m,-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right ) \left (-\frac {a (1-\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} \left (\frac {a (1+\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}}{a^3 d (1-m)}+\frac {b^2 e \operatorname {AppellF1}\left (2-m,\frac {1-m}{2},\frac {1-m}{2},3-m,-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right ) \left (-\frac {a (1-\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} \left (\frac {a (1+\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}}{a^3 d (2-m) (b+a \cos (c+d x))}+\frac {\cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{a^2 d e (1+m) \sqrt {\cos ^2(c+d x)}} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(1433\) vs. \(2(405)=810\).
Time = 11.66 (sec) , antiderivative size = 1433, normalized size of antiderivative = 3.54 \[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^2} \, dx=-\frac {4 b \operatorname {AppellF1}\left (\frac {1+m}{2},m,1,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) (b+a \cos (c+d x)) \sec ^2(c+d x) (e \sin (c+d x))^m \tan \left (\frac {1}{2} (c+d x)\right )}{a^2 d (a+b \sec (c+d x))^2 \left (\operatorname {AppellF1}\left (\frac {1+m}{2},m,1,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )+2 m \operatorname {AppellF1}\left (\frac {1+m}{2},m,1,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \cot (c+d x) \tan \left (\frac {1}{2} (c+d x)\right )+2 m \operatorname {AppellF1}\left (\frac {1+m}{2},m,1,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )-\frac {2 (1+m) \left ((-a+b) \operatorname {AppellF1}\left (\frac {3+m}{2},m,2,\frac {5+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+(a+b) m \operatorname {AppellF1}\left (\frac {3+m}{2},1+m,1,\frac {5+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{(a+b) (3+m)}\right )}+\frac {2 b^2 \left ((a+b) \operatorname {AppellF1}\left (\frac {1+m}{2},m,1,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )-2 a \operatorname {AppellF1}\left (\frac {1+m}{2},m,2,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right ) \sec ^2(c+d x) (e \sin (c+d x))^m \tan \left (\frac {1}{2} (c+d x)\right )}{a^2 d (a+b \sec (c+d x))^2 \left (\left ((a+b) \operatorname {AppellF1}\left (\frac {1+m}{2},m,1,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )-2 a \operatorname {AppellF1}\left (\frac {1+m}{2},m,2,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )+2 m \left ((a+b) \operatorname {AppellF1}\left (\frac {1+m}{2},m,1,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )-2 a \operatorname {AppellF1}\left (\frac {1+m}{2},m,2,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right ) \cot (c+d x) \tan \left (\frac {1}{2} (c+d x)\right )+2 m \left ((a+b) \operatorname {AppellF1}\left (\frac {1+m}{2},m,1,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )-2 a \operatorname {AppellF1}\left (\frac {1+m}{2},m,2,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )-\frac {2 (1+m) \left (\left (-a^2+b^2\right ) \operatorname {AppellF1}\left (\frac {3+m}{2},m,2,\frac {5+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+4 a (a-b) \operatorname {AppellF1}\left (\frac {3+m}{2},m,3,\frac {5+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+(a+b) m \left ((a+b) \operatorname {AppellF1}\left (\frac {3+m}{2},1+m,1,\frac {5+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )-2 a \operatorname {AppellF1}\left (\frac {3+m}{2},1+m,2,\frac {5+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{(a+b) (3+m)}\right )}-\frac {(b+a \cos (c+d x))^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3}{2},\cos ^2(c+d x)\right ) (e \sin (c+d x))^m \sin ^2(c+d x)^{\frac {1}{2} (-1-m)} \tan (c+d x)}{a^2 d (a+b \sec (c+d x))^2} \]
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\[\int \frac {\left (e \sin \left (d x +c \right )\right )^{m}}{\left (a +b \sec \left (d x +c \right )\right )^{2}}d x\]
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\[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
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\[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\left (e \sin {\left (c + d x \right )}\right )^{m}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]
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\[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
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\[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (e\,\sin \left (c+d\,x\right )\right )}^m}{{\left (b+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]
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